Brian Bi
$\DeclareMathOperator{\Ind}{Ind}$

Remark on prelude: The text claims that it's easy to see that $$U_\lambda$$ can be alternatively defined as $$\mathbb{C}[S_n]a_\lambda$$. I'm not sure it's quite so trivial. An element of $$\Ind^{S_n}_{P_\lambda} \mathbb{C}$$ is a function $$f : S_n \to \mathbb{C}$$ subject to the restriction that $$f(hg) = hf(g) = f(g)$$ for each $$g \in S_n, h \in P_\lambda$$; that is, $$f$$ is a constant when restricted to any right $$P_\lambda$$-coset of $$S_n$$. There's also a natural way to view $$\mathbb{C}[S_n]a_\lambda$$ as a set of functions from $$S_n$$ to $$\mathbb{C}$$, namely, by treating $$\sum_{g \in S_n} c_g g$$ as the function that assigns to each element $$g$$ the scalar $$c_g$$. However, such functions will be constant on each left coset and not necessarily on each right coset.

Observe, however, that by Problem 5.10.2(d), $$U_\lambda \cong \mathbb{C}[S_n] \otimes_{P_\lambda} \mathbb{C}$$, and $$\mathbb{C}[S_n]a_\lambda$$ is also isomorphic to $$\mathbb{C}[S_n] \otimes_{P_\lambda} \mathbb{C}$$ when $$x \otimes k \in \mathbb{C}[S_n] \otimes_{P_\lambda} \mathbb{C}$$ is identified with $$kxa_\lambda \in \mathbb{C}[S_n]a_\lambda$$. Therefore $$U_\lambda$$ and $$\mathbb{C}[S_n]a_\lambda$$ are isomorphic to each other. (The natural isomorphism defined in Problem 5.10.2(d) involves taking inverses, which is why it can be used to flip from right cosets to left cosets, so to speak.)

Remarks on proof of Theorem 5.14.3: The proof is very terse. Although there is nothing here that is specifically about representation theory, since I did the work of figuring out the details, I figured I'd share them.

Etingof states that for $$g \in C_{\mathbf{i}}$$, we have $Z_g \cong \prod_m S_{i_m} \ltimes (\mathbb{Z}/m\mathbb{Z})^{i_m}$ where $$Z_g$$ is the centralizer of $$g$$. This is based on the fact that conjugation in the symmetric group is relabelling of elements. Thus, $$Z_g$$ is the set of permutations $$h$$ such that if $$h$$ is used to relabel the elements of $$g$$, we get $$g$$ back. A necessary condition is that each element $$j \in \{1, \ldots, n\}$$ must be relabelled to an element that occurs in a cycle of the same length as $$j$$ in $$g$$. $$Z_g$$ is therefore a direct product of subgroups $$Z_{gm}$$ each of which permutes only those $$j$$ that occur in one of the cycles of length $$m$$ in $$g$$. For each $$j$$, once we have chosen the image $$h(j)$$, the element $$g^k(j)$$ must be sent by $$h$$ to $$g^k(h(j))$$ in order to get back the original $$g$$. That is, each cycle of length $$m$$ in $$g$$ must be mapped by $$h$$ in its entirety to one of the cycles of length $$m$$ in $$g$$ (possibly itself). Thus, each element of $$Z_{gm}$$ can be viewed as a permutation of the cycles themselves followed by a relabelling of each cycle mapping it to itself. The relabellings of an individual cycle $$c$$ mapping it to itself are simply $$\langle c \rangle$$. So we have $$Z_{gm} = (\mathbb{Z}/(m\mathbb{Z}))^{i_m} S^{gm}$$ where $$S^{gm}$$ is a subgroup of $$S_n$$ that maps each cycle of $$g$$ of length $$m$$ to some cycle of $$g$$. The two subgroups in this product have trivial intersection, and the first factor $$(\mathbb{Z}/(m\mathbb{Z}))^{i_m}$$ is a normal subgroup of $$Z_{gm}$$ since each element of $$S^{gm}$$ simply permutes its factors. We therefore have $$Z_{gm} \cong S^{gm} \ltimes (\mathbb{Z}/(m\mathbb{Z}))^{i_m}$$, a semidirect product. Finally, $$S^{gm}$$ is itself isomorphic to $$S_{i_m}$$.

Next, Etingof gives a formula for $$|C_{\bf i} \cap P_\lambda|$$, which involves summing over collections $$r_{jm}$$ specifying the number of cycles of length $$m$$ in the set of $$\lambda_j$$ elements corresponding to row $$j$$ of the tableau. For each such collection $$r_{jm}$$, we have to compute the number of ways to divide up $$\lambda_j$$ distinguishable elements into $$r_{jm}$$ cycles of length $$m$$ for each $$m$$. Denote this by $$A_{jm}$$. Then $A_{jm} = B_{jm} \times \left(\prod_m C_{jm}\right) \times \left(\prod_m D_{jm}\right)$ where $$B_{jm}$$ is the number of ways to partition $$\lambda_j$$ distinguishable elements into an ordered collection of subsets of $$mr_{jm}$$ elements for each $$m$$, representing the elements that will belong to cycles of length $$m$$; $$C_{jm}$$ is the number of ways to divide $$mr_{jm}$$ distinguishable elements into an unordered collection consisting of $$r_{jm}$$ subsets of $$m$$ elements each; and $$D_{jm}$$ is the number of ways to place $$mr_{jm}$$ distinguishable elements already so partitioned into $$r_{jm}$$ cycles of length $$m$$, each element forming a cycle together with the other elements in its subset. We have \begin{gather*} B_{jm} = \frac{\lambda_j!}{\prod_m (mr_{jm})!} \\ C_{jm} = \frac{1}{r_{jm}!}\frac{(mr_{jm})!}{(m!)^{r_{jm}}} \\ D_{jm} = (m-1)!^{r_{jm}} \end{gather*} where for $$B_{jm}$$ and $$C_{jm}$$ we have used the multinomial coefficients, and for $$D_{jm}$$ we used the fact that there are $$(m-1)!$$ $$m$$-cycles on a set of size $$m$$. The formula stated in the text for $$A_{jm}$$ then follows.

The subsequent formula for $$\chi_{U_\lambda}(C_{\bf i})$$ is obtained after a bit of algebra. The appearance of the product in the denominator of the expression for $$|C_{\bf i} \cap P_\lambda|$$ should not distract us from the fact that we can exchange the order of the two product operations.

Finally, to get the desired coefficient, we have the count the number of different ways that we can select one term from each of the factors in the product at the end, such that the product of the selected terms is $$x^\lambda$$. As Etingof points out, this can be done by considering each set of $$r_{jm}$$'s, stipulating that $$x_j^m$$ is taken $$r_{jm}$$ times, then summing over all possible $$r$$. Once we know that $$x_j^m$$ is taken $$r_{jm}$$ times, we have to select, for each $$j, m$$, a subset of the $$i_m$$ factors containing powers of $$m$$ of cardinality $$r_{jm}$$; this is the subset for which we pick $$x_j^m$$ from the corresponding factor. As we are picking $$x_1$$ $$r_{1m}$$ times, and so on from distinguishable factors, the total number of ways to select one term from each of the $$i_m$$ factors consistently with $$r_{jm}$$ is the multinomial coefficient $$\frac{i_m!}{\prod_j r_{jm}!}$$, and with $$r_{jm}$$ known, we can treat the $$x_j^m$$ with different $$m$$ independently. Thus the desired coefficient is $$\sum_r$$ of $$\prod_m$$ of the multinomial coefficient, just like in $$\chi_{U_\lambda}(C_{\bf i})$$. (Maybe this last paragraph just made you more confused. In that case, ignore it and try to work out for yourself how to compute the desired coefficient by summing over $$r$$. It will probably make sense then.)